# Isomorphism between groups...

• Nov 4th 2010, 09:58 PM
matt.qmar
Isomorphism between groups...
Hello!

I am trying to show that $\displaystyle \alpha$ is an ismorphism,
$\displaystyle \alpha:U(st) \rightarrow U(s) \oplus U(t)$
defined by $\displaystyle \alpha(x) = (x\ mod\ s, x\ mod\ t)$ where $\displaystyle x \in U(st)$
where U(n) is the multiplicative group of units (refresher: numbers relativly prime to n)
and $\displaystyle \oplus$ is the external direct product.

so clearly $\displaystyle \alpha$ is well defined
I have shown it is 1-1
So my problem is...
I am having trouble showing is it onto (to show to that it is a bijection)
so let $\displaystyle b = (y_1,y_2) \in U(s) \oplus U(t)$. So b is an arbitrary element of $\displaystyle U(s) \oplus U(t)$.

so what we need to do is find some $\displaystyle x \in U(st)$ such that
$\displaystyle \alpha(x) = b = (y_1,y_2)$ (we must find a x that maps to b... defn of onto, right?)

Keep in mind that we know that $\displaystyle gcd(s, y_1) = gcd (t, y_2) = 1$ since $\displaystyle gcd(s,t) = 1$

--------
So I thought maybe $\displaystyle x = y_1t +y_2s$?

then by the definition of alpha,

$\displaystyle \alpha(x) = ((y_1t +y_2s)\ mod\ s, (y_1t +y_2s)\ mod\ t) = ((y_1t)\ mod \ s, (y_2s)\ mod\ t)$
Now we just need to get rid of the t in the first component and s in the second component to map to that arbitrary element $\displaystyle b = (y_1, y_2) \in U(s) \oplus U(t)$

Yikes sorry for all the description. All help would be much appriciated, this actually stumped our class for a few minutes at the end of class today!

• Nov 4th 2010, 11:05 PM
roninpro
You can cop out a little bit by noting that your map is between finite sets. Since the map is injective, it is automatically surjective.
• Nov 5th 2010, 12:52 AM
Swlabr
Quote:

Originally Posted by roninpro
You can cop out a little bit by noting that your map is between finite sets. Since the map is injective, it is automatically surjective.

He'd also have to note that the sets are of the same size!
• Nov 5th 2010, 08:11 AM
matt.qmar
which they are!
$\displaystyle \phi(mn) = \phi(m)\phi(n)$ for Euler's phi function (which counts the number of elements in U(n).

thanks alot! I had totally forgotten that handy tool.

just for interests sake, still wouldn't mind to find an element mapping to b though!
• Nov 5th 2010, 08:30 AM
Drexel28
Quote:

Originally Posted by matt.qmar
which they are!
$\displaystyle \phi(mn) = \phi(m)\phi(n)$ for Euler's phi function (which counts the number of elements in U(n).

thanks alot! I had totally forgotten that handy tool.

just for interests sake, still wouldn't mind to find an element mapping to b though!

Don't you need to assume that $\displaystyle (s,t)=1$?
• Nov 5th 2010, 09:02 AM
matt.qmar
Yes, only holds for coprimes. But in this case, $\displaystyle gcd(s,t)=1$